How can infinity be an accumulation point?

Question

Let $(a_n)_{n∈\mathbb N}$ be a set in $\mathbb R$. For every $r ∈ \mathbb R$ exits an accumulation point $b ∈ R \cup \{{−\infty} \}$ of $(a_n)_{n∈\mathbb N}$ with $ b < r$ . ($\pm \infty$ are allowed as accumulation points)

Show, that $-\infty$ is an accumulation point of $(a_n)_{n∈\mathbb N}$.

I can't wrap my head around this: An accumulation point is a point of a set, which in every yet so small neighborhood (of itself) contains infinitely many points of the set, right?

If (in my case) $-\infty$ is an accumulation point, then we could find a point of the set, that is finitely distant from infintity?!

Answer 1

Which points of a metric (or topological) space are accumulation points depends on which metric (or topology) you are using. In the case of $\mathbb{R}\cup\{-\infty,\infty\}$, the usual metric $d(x,y)=\vert x-y\vert$ is not going to work, since it is not a metric on this space.

Since you have tagged the question as analysis rather than topology, I'll assume that you don't want to work in an explicitly topological framework. So instead, we use a sort of work around:

A sequence $\langle x_n\vert n\in\mathbb{N}\rangle$ converges to $\infty$ if for every $M$ there is a natural number $N$ such that $$\forall n>N \, (x_{n}>M).$$

In this case, the idea of getting "close to infinity" is therefore expressed as "gets arbitrarily large".

Similarly:

$\infty$ is an accumulation point of the sequence if for every real $M$ there is some $n$ such that $M<x_{n}<\infty$.

Note that if the sequence does not take the value $\infty$, this is equivalent to the sequence being unbounded above. Note also that for any $M$, there must in fact be infinitely many such $n$s.

Convergence to $-\infty$, and the property of having $-\infty$ as an accumulation point are defined analogously.

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So, you have a sequence $\langle a_{n}\vert n\in\mathbb{N}\rangle$ which has arbitrarily small accumulation points. To show that $-\infty$ is also an accumulation point, you just need to show that it is unbounded below. But this is straightforward: for every $M<0$, there is an accumulation point $b<(M-1)$, and so the sequence contains a term below $b+1$ (or $b+\varepsilon$ if you prefer, so long as $\varepsilon>0$). Thus $M$ is not a lower bound for the sequence. Since $M$ is arbitrary, the sequence has no lower bounds, and therefore has $-\infty$ as an accumulation point.

Answer 2

This means that, for any n and x, there are at least n points greater than x.

Just make n and x as large as you want.

For example, consider the positive integers.

Answer 3

We use it in two different ways, that result in the same thing. Either you are working in the extended reals, $\Bbb R \cup \{-\infty,\infty\}$ and define all intervals of the form $[-\infty,x)$ and $(y,\infty]$ to be open. Then $-\infty$ is an accumulation point of the sequence iff every open set containing $-\infty$ contains at least one element of the sequence. The elements do not have to be "a finite distance" from $-\infty$, they just need to be "close" in the sense of being in open sets. That means whatever number I give you, there are elements of the sequence that are more negative.

The other approach is to continue working in $\Bbb R$ and to define "having an accumulation point at $-\infty$" (even though $-\infty$ is not part of our set) as having elements that are more negative than any value I give you.

Answer 4

Here's the thing: when you add the points $\pm \infty$ into the real line, your space is no longer a metric space.

In particular, when we consider two points $x,y \in (-\infty,\infty)$, it makes sense to talk about the "distance between them", i.e. $|x - y|$. In this case, it makes sense to talk about decreasing the "size of a neighborhood" because each point in a neighborhood about $x$ has a distance from $x$, which we can say is less than some radius $r$.

When we extend the line by adding $\pm \infty$, it no longer makes sense to talk about the distance of any two points. In particular, we would have to say that every $x \neq \infty$ is "infinitely far away" from infinity, so no value of $|x - \infty|$ makes sense.

In order to deal with this, we extend our definition of open sets, and our definition of an accumulation point to handle this space which is only a "topological" space since it is no longer a metric space. We say that an open neighborhood of $\infty$ is a set of the form $(a,\infty]$, and that $[-\infty,a)$ is an open neighborhood of $- \infty$.

We can say that any point $x$ in our metric space is an accumulation point of a set $S$ if every neighborhood $U$ of $x$ contains a point in $S$ other than $x$. So, $\infty$ is an accumulation point of $S$ if $(a,\infty]$ contains a (finite) member of $S$ for each $a \in \mathbb{R}$.

Note that applying this definition to a finite $x \in \mathbb{R}$ gives you the usual notion of an accumulation point, since each neighborhood of radius $\frac 1n$ must contain a point in $S$ other than $x$.

Answer 5

You can try setting $b_n=\arctan a_n$ because $f:\mathbb R \to \left(-\frac {\pi}2,\frac {\pi}2\right)$ is a continuous function which has a continuous inverse. Everything then becomes nicely finite. This is a bit of a cheat, but is one way you might be able to visualise what is going on.